Chap 4 Heat Transfer (PART 2)
Transcript of Chap 4 Heat Transfer (PART 2)
BKF2422 HEAT TRANSFER
Chapter 2
Principles of steady-state heat transfer in conduction
TOPIC OUTCOMES
It is expected that students will be able to:
• Solve problems using steady-state conduction principles for one-dimensional solid conduction heat transfer in parallel and series
• Calculate overall heat transfer coefficient to solve problems related to combined conduction and convection heat transfer mechanism.
• Solve the problem related to internal heat generation and determine the critical thickness and of insulation for a cylinder
• Apply shape factor to estimate the multidimensional heat transfer
CONTENTS• One Dimensional Conduction Heat
Transfer– Conduction Through a Plane Wall
– Conduction Through Solids In Series
– Conduction Through Solids In Parallel
– Conduction Through a Hollow Cylinder
– Conduction Through a Multilayer Cylinders
– Conduction Through a Hollow Sphere
• Combined Conduction and Convection and Overall Heat Transfer Coefficient
CONDUCTION: FOURIER’S LAW• Flux of conduction heat transfer can be calculated by
Fourier’s LawFourier’s Law
qx : heat-transfer rate in the x direction (SI: W or J/s; cgs: cal/s; Eng.: btu/h)
A : cross sectional are normal to the heat flow (m2)k : thermal conductivity ( SI: W/m. K; cgs: cal/s. cm. °C;
Eng.: btu/h. °F. ft )dT/dx : temperature different in the x direction
• The minus sign is required in Fourier’s equation because the heat transfer is positive in the direction from initial point 1 to the final point 2. Since the T1 > T2 (heat is transport from high temperature to lower temperature region), minus sign is needed to make the value of heat rate positive.
dx
dTk
A
qx
HEAT TRANSFER – CONDUCTION
• For steady state, the equation can be integrated,
This equation is basically a matter of putting in values to solve.
21
12
2
1
2
1
TTxx
k
A
q
dTkdxA
q
x
T
T
x
x
x
HEAT TRANSFER – CONDUCTIONConduction Through a Plane Wall.
The temperature various linearly with distance.
0 ΔxΔx Distance,x (m)
T1
T2
Temperature, (K)
T2
T1
q
R
TT
kAx
TTq 2121
EXERCISE 1
Calculate the heat loss per m2 of surface area for an insulating wall composed of 25.4 mm thick fiber insulating board, where the inside temperature is 352.7 K and the outside temperature is 297.1K.
From Table A.3 (pg 599), thermal conductivity for fiber insulating board is 0.048 W/m.K.
1 2
2 1
2
( ) 0.048(352.7 297.1)
0.0254
105.1 /
k T Tq
A x x
qW m
A
Conduction Through Solids In Series.
T1 A B C
q T2 T3
DxA DxB DxC T4
• The rate of heat transfer,
where,
CRRR
TT
R
TT
R
TT
R
TTq
BACBA
41433221
Ak
xR
A
AA
Ak
xR
B
BB
Ak
xR
C
CC
HEAT TRANSFER – CONDUCTION
EXERCISE 2
A cold storage room is constructed of an inner layer of 12.7 mm of pine, a middle layer of 101.6 mm of cork board and an outer layer of 76.2 concrete. The wall surface temperature is 255.4 K inside the cold room and 297.1 K at the outside surface of concrete. The conductivites for pine, 0.151; cork board, 0.0433; and concrete, 0.762 W/m.K. Calculate the heat loss in W for 1 m2 and the temperature at the interface between wood and cork board.
Answer: (-16.48 W, 256.79 K)
SOLUTION
W
K
Ak
xR
W
K
Ak
xR
W
K
Ak
xR
mxmxmx
kkk
mAKTKT
C
CC
B
BB
A
AA
CBA
CBA
??
?
346.2)1(043.0
1016.0
0841.0)1(151.0
0127.0
0762.0,1016.0,0127.0
762.0,043.0,151.0
1,1.297,4.255 2
41
HEAT TRANSFER – CONDUCTION
• Conduction Through Solids In Parallel
A
B
C
D
E
F
G
1 2 2 3 2 3 2 3( ) ( ) ( ) ( ) .......
T A B C D E F G
C CA A B B D DA
A B C D
q q q q q q q q
k Ak A k A k Aq T T T T T T T T
x x x x
HEAT TRANSFER – CONDUCTION
Conduction Through A Hollow Cylinder.T2 L
r2
q
r1
• The cross-sectional area normal to the heat flow is, A =2prL.
• The rate of heat transfer,
dr
dTk
A
q
CONDUCTION THROUGH A HOLLOW CYLINDER
2
1
2
12
r
r
T
TdTk
r
dr
L
q
dr
dTk
A
q
p
12
12
12
12
12
12
ln
)(2
22ln
22
ln
rr
rrL
LrLr
LrLr
AA
AAAlm
p
pp
pp
12
21
rr
TTkAq lm
or
21
12ln
2TT
rr
Lkq
p
Where:
kL
rr
kA
rrR
R
TT
kArr
TTq
lm
lm
p2
ln 1212
21
12
21
EXERCISE 3A thick wall cylindrical tubing of hard rubber having
andinside radius of 5 mm and an outside radius of 20
mm is being used as a temporary cooling coil in a bath. Ice
water is flowing rapidly inside, and the inside wall
temperature is 274.9 K. The outside temperature is at 297.1 K. A
total of 14.65 W must be removed from the bath by the
cooling coil. How many m tubing are needed?(k=1.15
W/m.K)
1 2
2 1
2( )
ln( / )
20.151 (274.9 297.1)
ln(0.02 / 0.005)
15.2 /
14.650.964
15.2 /
Lq k T T
r r
Lq x
qW m
L
Wlength m
W m
p
p
HEAT TRANSFER – CONDUCTION
Conduction Through a Multilayer Cylinders.
Example, heat is being transferred through the walls of an insulated pipe.
T1
T2
T3
T4
r1
r2r3
r4q
A
B
C
HEAT TRANSFER – CONDUCTION
• At steady-state, the heat-transfer rate q, be the same for each layer.
• The rate of heat transfer,
where,
CBACBA RRR
TT
R
TT
R
TT
R
TTq
41433221
Lk
rrR
A
Ap2
ln 12
Lk
rrR
B
Bp2
ln 23
Lk
rrR
C
Cp2
ln 34
EXERCISE 4
A thick walled tube of stainless steel (A) having a k = 21.63 W/m.K with dimensions of 0.0254 m ID and 0.0508 m OD is covered with a 0.0254 thick layer of insulation (B), k = 0.2423 W/m.K. The inside wall temperature of the pipe is 811 K and the outside is at 310.8 K. For a 0.305 m length pipe, calculate the heat loss and also the temperature at the interface between the metal and the insulation.
Answer: (331.7 W, 805.5 K)
SOLUTION
A
BA
lmA
A
Alm
R
TTq
WRR
TTq
WKAk
rrR
mAA
AAA
AmLrA
mrmr
21
31
12
2
12
12
2
2
11
21
7.331
/01673.0)0351.0(63.21
0127.0
0351.00243.0/0487.0ln
0243.00487.0
/ln
,0243.00127.0305.022
......0254.02
0508.0,0127.0
2
0254.0
pp
HEAT TRANSFER – CONDUCTION
Conduction Through a Hollow Sphere
T2 r2
q r1 T1
• The cross-sectional area normal to the heat flow is, A = 4pr2.
• The rate of heat transfer,dr
dTk
A
q
R
TT
krr
TTTT
rr
kq
dTkr
drq r
r
T
T
21
21
2121
21
2
41111
4
4
2
1
2
1
p
p
p
HEAT TRANSFER – CONDUCTION
HEAT TRANSFER – CONDUCTION
Temperature Profile for Heat Transfer By Convection From One Fluid To Another.
film filmMetal wall
Warm liquid A
Cold fluid B
T6
T5
T4
T3
T2 T1
q
HEAT TRANSFER – CONDUCTIONRegion,T1 – T2: turbulent fluid flow. Mainly convective heat transfer.T2 – T3: velocity gradient very steep. No turbulent flow,
(i.e. only laminar). Mainly conductive heat transfer.T3 – T4: conductive heat transfer.T4 – T5: no turbulent in film, mainly conductive heat transferT5 – T6: turbulent flow, conductive heat transfer.T1 – T2 and T5 – T6 : different are small.
Convective coefficient for heat transfer through a fluid:q = hA(T – Tw)
where,h = convective heat transfer coefficient.T = average temperature in fluid.Tw = temperature of wall in contact.
HEAT TRANSFER – CONDUCTION
Combined Convection and Conduction and Overall Coefficients.
• Heat flow with convective boundaries: plane wall
oAi
oi
oAAi
oi
RRR
TT
AhAkxAh
TTq
11
oi
oAAi
oi TTUAhkxh
TTAq
11
TiT1
T2
To
q
hi
ho
xA
HEAT TRANSFER – CONDUCTION
Heat flow with convective boundaries: cylindrical wall with insulation.
T1
T2
T3
T4
rO
ri
r1ho
hi
A
B
oAAoioiio
o
ooiAAiioi
i
ooii
ooAAioii
hAkArrhAAU
hAAAkArrhU
R
TTTTAUTTAUq
R
TT
AhAkrrAh
TTq
lm
lm
lm
1
1
1
1
11
414141
4141
Similarly,
Overall heat transfer coefficient,
where,
and
HEAT TRANSFER
• Other way we can used,
where,
414141 TTAUTTAU
RRRR
TTq ooii
oBAi
Lk
rrR
A
iA
p2
ln 1
Lk
rrR
B
oB
p2
ln 1
iiii
iAhhLr
R1
2
1
p oooo
oAhhLr
R1
2
1
p
A thick-walled tube of stainless steel (A)having a k = 21.63
W/m.k with dimensions of 0.0254m ID and0.0508m OD is
covered with a 0.0254m layer of asbestos (B)insulation, k =
0.2423 W/m.k. The inside wall temperature ofthe pipe is 811K
and the outside surface of the insulation is at310.8K. For a
0.305m length of pipe, calculate the heat lossand also the
temperature at the interface between themetal and the
BA RR
TTq
31
The resistances are
K/W 01673.0
)305.0)(63.21(2
)0127.0
0254.0ln(
2
)ln(
2
)ln( 1/21/2
p
pp Lk
dd
Lk
rrR
AA
A
K/W 493.1
)305.0.0)(2423.0(2
)0508.0
1016.0ln(
2
)ln(
2
)ln( 1/21/2
p
pp Lk
dd
Lk
rrR
BB
B
The heat transfer rate is
BA RR
TTq
31
W7.331
493.101673.0
8.310811
q
K 5.805
01673.0
8117.331
2
2
21
T
T
R
TTq
A
CRITICAL THICKNESS OF INSULATION FOR A CYLINDER
• If outer radius < rcr: adding more insulation will increase heat transfer rate
• If outer radius > rcr: adding more insulation will decrease heat transfer rate
h
kr cr )( 2
)( 120 TTAhq
02
12
01
1/ln
)(2
hrk
rr
TTLq
p
With insulation:
EXAMPLE
• An electric wire having a diameter of 1.5 mm covered with a plastic insulation (thickness = 2.5mm) is exposed to the air at 300K and ho = 20 W/m2.K. It is assumed that the wire surface temperature is constant at 400K and is not affected by the covering.
• A) calculate the value of the critical radius• B) calculate the heat loss per (m) of wire length with no
insulation• C) repeat (b) for insulation being present
a) 20 mm b) 9.42W c) 32.98 W
Convection: Heat transfer using movement of fluids.
Heat transfer is considered as convection with the presence of bulk fluid motion. Fluid motion enhances heat transfer where the higher the fluid velocity, the higher the rate of heat transfer.
2 main classification of convective heat transfer;
1. Forced Convection : fluid flow by pressure differences, a pump, a fan and so on
2. Natural Convection: motion of fluid results from the density changes in heat transfer
CONVECTION HEAT TRANSFER
The rate of heat transfer :
Tw = 80 oC
To = 30 oCq
q Ah(Tw To)The convection coefficient is a measure of how effective a fluid is at carrying heat to and away from the surface.
h = heat transfer coefficient
(W/m2.K)
A= surface area (m2)
Fluid flow
CONVECTION HEAT TRANSFER
Metal wall
Warm fluid A
Cold fluid B
q
Turbulence absent
T2
T3 Turbulence region
T1
q = hA (T-Tw)
FKKSA
FORCED CONVECTION INSIDE PIPES
Forced convection – fluid forced to flow by pressure differences
Types of fluid, laminar or turbulent
– great effect on heat-transfer coefficient
More turbulent– greater heat-transfer coefficient
Reynolds number, NRe
NReD
wherev = velocity of fluid (m/s)
= viscosity of fluid (Pa.s)
= density of fluid (kg/m3)
D = diameter of pipe (m)
FKKSA
FORCED CONVECTION
where
= viscosity of fluid (Pa.s)
= density of fluid (kg/m3)
k = thermal conductivity of fluid (W/m.K)
cP = heat capacity of fluid (J/kg.K)
h = heat transfer coefficient (W/m2.K)
D = diameter of pipe (m)
Prandtl number, NPr
Dimensionless numbers:
Nusselt number, NNu
k
μc
ρck
ρμ
N P
P
Pr
khDN
Nu
FKKSA
LAMINAR FLOW INSIDE HORIZONTAL PIPE
whereD = inside diameter of pipe (m)
L = length of pipe (m)
b = viscosity of fluid at bulk temperature (Pa.s)
w = viscosity of fluid at wall temperature (Pa.s)
ha = average heat transfer coefficient (W/m2.K)
NNu
a
haD
k1.86 N
ReN
Pr
DL
13
b
w
0.14
NRe 2100 & NReNP r 100 :L
D
All physical properties at except w2
biT
boT
mean bT
q = haA∆Ta where 2
boT
wT
biT
wT
aΔT
100
2100
PrRe
Re
L
DNN
N
Limitations
FKKSA
TURBULENT FLOW INSIDE HORIZONTAL PIPE
where
NNu
hLD
k0.027 N
Re
0.8N
Pr
13 b
w
0.14
NRe 6000 , 0.7 ≤ NP r ≤ 16000 & 60:DL
Rate of heat transfer is greater
cP = heat capacity of fluid (J/kg.K)
D = inside diameter of pipe (m)
k = thermal conductivity of fluid (W/m.K)
b = viscosity of fluid at bulk average temperature (Pa.s)
hL = heat transfer coefficient based on the log mean driving force
∆Tlm (W/m2.K)
w = viscosity of fluid at wall temperature (Pa.s)
Many industrial heat transfer processes in the turbulent region
60
160007.0
6000
Pr
Re
D
L
L
DN
N
Limitations
EXAMPLE 4.5-1 Page 262: Heating of Air inTurbulent Flow
Air at 206.8 kPa and an average of 477.6 K is being heated as
it flows through a tube of 25.4mm inside diameter at velocity
of 7.62 m/s. The heating medium of 488.7 K steam
condensing on the outside of the tube. Since the heat-transfer
coefficient of condensing steam is several thousand W/m2.K
and the resistance of the metal wall is very small, it will be
assumed that the surface wall temperature of the metal in
contact with the air is 488.7 K. Calculate the heat-transfer
coefficient for an L/D > 60 and also the heat-transfer flux
q/A.
boT
K 7.488 Steam, wT
steamo hh
L
mm 5.42air
biTkPa 8.206
m/s 62.7
K 6.477
P
v
Tave
3
Pr
5
kg/m 74.0
W/m03894.0
686.0
Pa.s 106.2
K 477.6 kPa, 101.32at A.3,Appendix From
k
N
TTP
b
bmave
Pa.s 1064.2
K 7.884at A.3,Appendix From
5
w
wT
3
kPa 8.206
21
1
2
1
212
kg/m 509.135.101
8.20674.0
on depend is
,For
TT
T
T
P
P
P & TRT
PM
RTPMRTV
mPM
RTM
mPVnRTPV
)6000(10122.1
106.2
)509.1)(62.7(104.25
4
5
3
Re
DN
.K W/m2.63
0264.0
0260.0686.010122.1027.0
03894.0
)104.25(
027.0
2
14.0
3
18.043
14.0
3
1
Pr
8.0
Re
L
L
w
bLNu
h
h
NNk
DhN
2 W/m1.701
6.4777.4882.63
bmwL TThA
q
flowrent countercur
flow parallel
hoThiT
hiT hoT
coT
coTciT
ciT
1T
1T
2T
2T
hiT
hiT
hoT
hoT
ciT
ciT
coT
coT
EXAMPLE 4.5-4 Page 268: Heat Transfer Area andLog Mean Temperature Difference
A heavy hydrocarbon oil which has a cpm = 2.30kJ/kg is
being cooled in a heat exchanger from 371.9 K to 349.7 K
and flows inside the tube at a rate of 3630 kg/h. A flow of
1450kg water/h enters at 288.6K for cooling and lows
outside the tube.
a) Calculate the water outlet temperature and heat-transfer
area if the overall Ui = 340 W/m2.K and the streams are
countercurrent
b) Repeat for parallel flow
flowrent countercur (a)
1T
2T
hiTK 1.397
K7.349hoT
K 6.288ciT
coT
kg/h 3630 oil, m
kg/h 1450 water, m
W51490
3600/)7.3499.371)(3.2(3630
hihohp TTcmq
?
.K W/m340
kJ/kg.K 187.4
kJ/kg.K 3.2
2
i
i
waterp
oilp
A
U
c
c
K 1.319
)6.288)(187.4(145051490
co
co
cicocp
T
T
TTcmq
K 9.56
8.52
1.61ln
8.521.61
ln2
1
21
T
T
TTTlm
2m 66.2
)9.56(34051490
i
i
lmii
A
A
TAUq
K 1.616.2887.349
K 8.521.3199.371
2
1
T
T
flow parallel (b)
K 3.836.2889.371
K 6.301.3197.349
2
1
T
T
K 7.52
6.30
3.83ln
6.303.83
ln2
1
21
T
T
TTTlm
2m 87.2
)7.52(34051490
i
i
lmii
A
A
TAUq
forces. driving peraturelarger tem
gives wscounterflo because occurs This w.counterflofor than arealarger a is This
•Radiation heat transfer is the transfer of heat by electromagnetic radiation•Occur in solid, liquid and gas•Not require heat transfer medium•Fastest energy transfer•Example: microwave, radar, cordless telephones
Absorptivity
• When thermal radiation (light waves) falls upon a body, part is absorbed, part is reflected into space and part is transmitted through the body.
• BLACK BODY – one that absorb all radiant energy and reflect none.
0reflectedn ty/fractioreflectivi
1.0absorbedn ty/fractioabsorptivi
• Kirchoff’s Law states at the same temperature T1
• For
• For a perfect black body with :
• Substances that have emissivity < than 1.0 are called gray bodies
1 body,Black
bodyblack ofpower emissive total
surface a ofpower emissive total ,Emissivity
BE
E
1 body,Gray
11
1 4TAq
4TAq